Calcium carbonate precipitate and the spectator ions in an aqueous solution for the following reaction.
Na2CO3(aq) + CaCl2(aq) –
CaCO3(s) + 2NaCl(aq)

13. calculate the initial concentration (before precipitation) of carbonate ions after the addition of each 0.05 ml of solution b to the 1.00 l beaker of solution a. divide the work among group members and write the answers in the table in model 3. assume the volume change as solution b is added is negligible. 14. notice the initial concentrations of zn2+ - and cu2+ in the table in model 3. a. explain how these were obtained from the data in model 2. b. as solution b is added and precipitates form, do these initial concentrations change? 15. use the data in model 2 to indicate the presence of precipitate (either znco3 or cuco3) after each 0.05 ml addition of solution b in model 3. 16. use the initial concentrations of carbonate ions and zinc ions to calculate the reaction quotient, qsp for the zinc carbonate scenarios in model 3. divide the work among group members and write the answers in the table in model 3. 17. use the initial concentrations of carbonate ion and copper(ii) ions to calculate the qsp for the copper(ii) carbonate scenarios in model 3. divide the work among group members and write the answers in the table in model 3.

13. a covalent bond between two atoms is likely to be polar if: a. one of the atoms is much more electronegative than the other. b. the two atoms are equally electronegative. c. the two atoms are of the same element. d. the bond is part of a tetrahedrally shaped molecule. e. one atom is an anion.

Answer the questions about this reaction: nai(aq) + cl2(g) → nacl(aq) + i2(g) write the oxidation and reduction half-reactions: oxidation half-reaction: reduction half-reaction: based on the table of relative strengths of oxidizing and reducing agents (b-18), would these reactants form these products? write the balanced equation: answer options: a. 0/na -> +1/na+1e- b. nai(aq) + cl2(g) → nacl(aq) + i2(g) c. +1/na+1e- -> 0 /na d. -1/2i -> 0/i2+2e- e. no f. 4nai(aq) + cl2(g) → 4nacl(aq) + i2(g) g. 2nai(aq) + cl2(g) → 2nacl(aq) + i2(g) h. 4nai(aq) + 2cl2(g) → 4nacl(aq) + 2i2(g) i. nai(aq) + cl2(g) → nacl(aq) + i2(g) j. 0/cl2+2e -> -1/2cl- k. yes